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Beliebt Rechnen >

y^'=y(2+y)t,y(0)=1

Lösung

y^{^{'}} = y (2 + y) t, y (0) = 1

Lösung

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

Schritte zur Lösung

y^{'} (t) = y (2 + y) t

Solve Bernoulli ODE:

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

Graph

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